The forum discussion centers on proving the inequality challenge VI for the sequence defined by $\alpha_n = \arctan n$. Participants confirm that for all integers $n \geq 1$, the condition $\alpha_{n+1} - \alpha_n < \frac{1}{n^2+n}$ holds true. The discussion highlights the mathematical rigor involved in establishing this inequality, with specific acknowledgment to participant Albert for his contributions to the proof.
PREREQUISITESMathematicians, students studying calculus, and anyone interested in advanced mathematical proofs and inequalities will benefit from this discussion.
anemone said:If $\alpha_n=arc\tan n$, prove that $\alpha_{n+1}-\alpha_n<\dfrac{1}{n^2+n}$ for $n=1,\,2,\,\cdots$.
Albert said:$tan\,\alpha_n=n$, and ,$tan\,\alpha_{n+1}=n+1 $ for $n=1,2,\,\cdots$
$tan(\,\alpha_{n+1}-\,\alpha_n)=\dfrac{1}{n^2+n+1}---(1)$
$\dfrac{1}{n^2+n}---(2)$
$tan\,\dfrac{1}{n^2+n}---(3)$
comare (1)(2)(3) and we prove it
(3)>(2)>(1) for n=1,2,...